Equilibrium of Cylinder with two liquids at either side

[ron_jay]

Homework Statement

(Please refer to the attachment given)

In the figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities $$2\rho$$ and $$3\rho$$ . The height h for the equilibrium of cylinder must be:

$$a) \frac{3R}{2}$$

$$b) R \sqrt{\frac{3}{2}}$$

$$c) R \sqrt{2}$$

$$d) R \sqrt{\frac{3}{4}}$$

Homework Equations

Basic Equations of hydrostatics

The Attempt at a Solution

This Problem is a little confusing. I think we have to consider the components of the pressures at various points on the cylinder, but I am not too sure how.Besides the above question, how would the cylinder move in the given state of inequilibrium?

 

[Doc Al]

Hint: Consider imaginary vertical walls just touching both sides of the cylinder. Now examine the forces acting on those walls.

 

[ogb p]

I would approach this problem by first drawing a free body diagram of the cylinder and analyzing the forces acting on it. The weight of the cylinder would act downward at its center of mass, while the buoyant forces of the two liquids would act upward at their respective contact points with the cylinder. The pressure at the bottom of the cylinder would be greater than the pressure at the top, due to the difference in densities of the two liquids.

To determine the height h at which the cylinder would be in equilibrium, I would use the equation for hydrostatic pressure: P = ρgh, where P is the pressure, ρ is the density, g is the acceleration due to gravity, and h is the height.

Since the cylinder is in equilibrium, the net force acting on it must be zero. Therefore, the upward buoyant forces must be equal to the downward weight of the cylinder. This can be expressed as:

F_b1 + F_b2 = mg

Where F_b1 and F_b2 are the buoyant forces of the two liquids, and m is the mass of the cylinder.

Using the equation for hydrostatic pressure, we can express the buoyant forces as:

F_b1 = ρ_1gh_1 and F_b2 = ρ_2gh_2

Substituting these into the equation above and rearranging, we get:

ρ_1gh_1 + ρ_2gh_2 = mg

Since we know that ρ_1 = 2ρ and ρ_2 = 3ρ, we can substitute these in and simplify to get:

2ρgh_1 + 3ρgh_2 = mg

Dividing both sides by ρg, we get:

2h_1 + 3h_2 = h

Since the cylinder is in contact with both liquids, the height of the cylinder must be equal to the sum of the heights of the two liquids, h = h_1 + h_2. Substituting this in, we get:

2h_1 + 3(h - h_1) = h

Simplifying, we get:

h_1 = h/2

Therefore, the height h for equilibrium of the cylinder must be h = 2h_1 = R√(3/2).

To address the second part of the question, the cylinder would not move in this state of equilibrium, as the